Re: juggling's combinatorics problem with siteswap

From: Rory Parle (rparle_at_soylentred.REMOVECAPS.net.INVALID)
Date: 02/09/05 

Date: Wed, 09 Feb 2005 20:29:19 +0000

Xah Lee wrote:
> Given a siteswape notatiion, is there a algorithm to derive:
>
> * the period number. (the period is the number of throws all colored
> balls return to their position.)

That's equal to the lowest common multiple of the individual orbit
periods, multiplied by two if it's odd.

> * the number of orbits in a pattern. (a orbit is the path a ball
> travels.)

- Let n0 = 0.
- Look at the number in position n0. Call it n1.
- Look at the number in position (n0 + n1) mod p (where p is the length
of the siteswap). Call that n2.
- Look at the number in position (n1 + n2) mod p. Call it n3.
- Continue in this vein until you get back to position 0. All of the
positions you've been to so far are part of the first orbit.
- Go to the next number not yet examined and repeat all the previous
steps starting from it (this time n0 is the position of that initial
number).
- Continue all of these steps until you've counted all the orbits.

> * whether the pattern is ambidextrous. (a pattern is ambidextrous if
> left and right hand play the same role. (not necessarily synchronized
> or lockstep.))

If the length of the siteswap (usually called period but different from
the period you asked about) is odd then it's ambidextrous. 

-- 
Rory Parle
http://www.netsoc.dit.ie/~jugsoc/

Relevant Pages

  • Re: Alternative proofs of the average theorem?
    ... Let the siteswap be the sequence a_i, where i goes from 0 to n-1. ... Let the index of this orbit be m = m), ... For example, if our siteswap is 453, then the permutation ... The ball (i.e. an orbit of the induced function f, ...
    (rec.juggling)
  • Re: A Rapture Of Siteswap (reprise)
    ... The colored period of a siteswap is simply the lowest common multiple ... of the sums of the throw heights of each orbit. ... and another orbit of a single ball goes from hand to hand. ...
    (rec.juggling)
  • Re: drawing a valid siteswap
    ... Split the 51 orbit into 6 segments, and do the same with the 3 orbit. ... The 5 is thrown first, so put it out of the hand by two segments, put the 3 out of the other hand by 1 segment, and put the 1 in the hand. ... You could of course split the orbits into 12 points, and you would find that all three balls peaked at the same time. ... not I end up drawing an accurate depiction of a siteswap. ...
    (rec.juggling)
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